How do I find next bit to change in a Gray code in constant time?

Question

I have a small 8-bit processor which has a N-to-M decoder on some output lines - eg, for the 5 to 32 bit case, I write 00101 and bit 5 changes state. The only interface to

Answer 1

You don't need to calculate the Gray codes and xor them, you can just use the counter itself, and then use a 256-element lookup table to count the number of trailing zeros. Like this:

unsigned char bit_change(unsigned char counter[4]) {
  static const unsigned char ones[] = {
    0,0,0,1,0,1,1,2,0,1,1,2,1,2,2,3,0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,
    0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
    0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
    1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,
    0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
    1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,
    1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,
    2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,    
  };

  unsigned char i;
  for (i = 0; i < 4; i++) {
    unsigned char x = counter[i];
    if (x) {
      x ^= x - 1;
      return 8 * i + ones[x];
    }
  }
}

If you unroll the loop, this is at most 2 adds, 1 xors, and 5 loads (but almost always less). If you don't have 256 bytes for the table, you could use the same strategy on nibbles.